Another brain teaser?

Asked by LostInParadise (32186

) February 12th, 2009

I believe I got this from the Ask Marilyn column in the Parade magazine newspaper insert.

A man is standing at the roulette wheel in a casino when he is approached by someone giving the following spiel: “Why should the casino be taking our money? Why not gamble between the two of us? Here is what I propose. Each of us selects a sequence of 3 color choices of red or black. The winner will be the one whose sequence first shows up on roulette wheel spinnings. This is of course totally random, but to show what a good sport I am, I will let you select your colors first.”

This is in fact a hussle. By going second and making the right color choices, the hussler always has a better than 50 percent chance of winning. Suppose that the offer is accepted and the first man chooses the sequence: red, red, red. What would the other man choose? What would be the chances of red, red, red being the winning sequence?

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4 Answers

Not sure on this one but the way I see it is as a red tile is not removed from the wheel each time it is hit the odds of hitting a red the second time is exactly the same as the first time and so on.
that chance is just under 50% because there are an equal number of black tiles plus a green tile.

If I was the hustler I would go for black, red, red as before his sequence of 3 reds so I would only need 2 reds to show up in a row while he is waiting for 3 (assuming that it does go green red red red).

kullervo (785

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Your intuition is spot on. Black, red, red is what the hustler would choose. The first person can win if the first three spins are all red. What are his chances otherwise?

LostInParadise (32186

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There are 37/38 tiles and unless he gets 3 reds in a row he has a 1 in 37/38 chance of winning as it would have to land on green before the 3 reds.

If they are negating the green and saying just out of red and blacks then he has no chance once a single black tile shows up. But as I said previous results do not influence future results so the odds are 50:50 of getting a red from the start and then 50:50 for the next one etc.

kullervo (785

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You got the main point. Once a single black shows up, there is no chance of the first person winning. So the he wins only if the first three times are all red, which happens with probability 1/8.

LostInParadise (32186

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